Let $f: (\mathbb{Q^{+}_r} \cup \{0\}) \rightarrow \mathbb{Z^+}$ by

$\begin{array}{cc}\Bigg \{&\begin{array}{cc} f(0)=1 \\ f(\frac{a}{b}) = a+b \end{array} \end{array}$

where $\mathbb{Q^{+}_r} = \big \{\frac{a}{b} \in \mathbb{Q} \big | a,b \in \mathbb{Z^+}, \frac{a}{b} \thinspace is \thinspace reduced \big \}$

Determine if f is surjective or injective.

I don't know how to go about this problem.

Also, if all but one of the branches of a piece-wise defined function is injective, does that necessarily mean that the function is not injective, or does it mean that f is not injective under the certain condition?

  • 1
    $\begingroup$ I am familiar with the terminology "a function that is defined piecewise" but not with "the piece of a function". Is this the restriction of fhe function to some subset of its domain? Have the "pieces" all the same codomain? Actually every partition of the domain induces a way to define a function piecewise. $\endgroup$ – drhab May 3 '15 at 15:24
  • $\begingroup$ I edited the question to be more precise in what I'm asking. $\endgroup$ – Benedict Voltaire May 3 '15 at 15:56

From e.g. $\frac14\neq\frac23$ combined with: $$f\left(\frac14\right)=1+4=2+3=f\left(\frac23\right)$$ it follows immedeately that $f$ is not injective.

If it comes to surjectivity then the question is: can we find for every $n\in\mathbb Z_{>1}$ some coprime pair $a,b\in\mathbb Z^+$ with $a+b=n$?

Yes we can: take $a=1$ and $b=n-1$. So the $f$ in your example is surjective.

If a function $f:X\rightarrow Y$ is injective then every restriction of $f$ to a subset $A\subset X$ is injective. So conversely if you can find a subset $A$ such that $f\upharpoonleft A$ is not injective then $f$ is not injective.

If $f:A\cup B\rightarrow Y$ is defined piecewise then - as argued above - it can only be injective if $f\upharpoonleft A$ and $f\upharpoonleft B$ are injective. However injectivity of $f\upharpoonleft A$ and $f\upharpoonleft B$ does not guarantee that $f$ is injective. It can happen that $f(a)=f(b)$ for $a\in A$ and $b\in B$ with $a\neq b$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.